Cow Contest

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M(1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 5
4 3
4 2
3 2
1 2
2 5

Sample Output

2
#include<stdio.h> 
#include<string.h>
/*
*	题解：
		如果该点与其他点都有路径相连，说明该点排名确定，否则不能 
*/
const int INF = 0x3f3f3f3f;
bool map[120][120];//判断是否有关系 

void floyd(int n){ //输入n个cows 
	for(int k = 1;k<=n;k++){
		for(int i = 1;i<=n;i++){
			for(int j = 1;j<=n;j++){
				if(map[i][k] && map[k][j]){
				//通过k看看能否使得i与j建立关系 
					map[i][j] = true;   //i->j有路，同时i在j的前面 
				}
			}
		}
	}
}

int main(){
	int n,m;//n个cows，m个关系 
	while(scanf("%d%d",&n,&m) != EOF){
		memset(map,false,sizeof(map));
		int u,v;//胜利方u，失败方v 
		for(int i = 0;i < m;i++){
			scanf("%d%d",&u,&v);
			map[u][v] = true;//map记录u和v是否有关系 
		}
		
		floyd(n);
		int cnt = 0;//记录有多少个点与其他所有点都有联系 
		for(int i = 1;i<=n;i++){
			bool flag = true;
			for(int j = 1;j<=n;j++){
				if(i!=j){//自己跟自己无需比较 
					if(!map[i][j] && !map[j][i]){ 
					//i不在j前面并且j不在i前面，说明关系不确定 
						flag = false;
						break;
					}
				}
			}
			if(flag){
				cnt++;
			}
		}
		printf("%d\n",cnt);
	}
	return 0;
}